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3 years ago

Calculated the gravitational potential energy of a 93.0 kg sky diver who is 550 m above the ground

1 answer:

4 0

Equation for Gravitation Potential Energy...
GPe = m·g·h
where m = mass in kg = 93.0 kg
g = force of gravity = 9.81 m/s²
<span>h = height in meters = 550m
</span>
GPe = mgh
GPe = (93)(9.81)(550)
GPe = 501781.5 Joules (J)
GPe = 5.02 x 10⁵ J (rounded & in scientific notation)

Therefore, the gravitational potential energy of a person of mass 93.0kg who is 550m above the ground is GPe = 501781.5 Joules (J)

Hope this helps!

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If the moment acting on the cross section is M=630N⋅m, determine the maximum bending stress in the beam. Express your answer to

-BARSIC- [3]

Answer:

2.17 Mpa

Explanation:

The location of neutral axis from the top will be

$\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm$

Moment of inertia from neutral axis will be given by $\frac {bd^{3}}{12}+ ay^{2}$

Therefore, moment of inertia will be

$\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}$

Bending stress at top= $\frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa$

Bending stress at bottom= $\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03$ Mpa

Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa

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3 years ago

Should you be worried if a n eighteen wheeler truck is riding close behind your car? Explain why or why not. Please help me D:

xz_007 [3.2K]

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3 years ago

Today's topic

babunello [35]

Answer:

While a body is said to be in motion if it changes its position with respect to immediate surroundings.

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2 years ago

A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.7 m/s rel

mafiozo [28]

Here it is given that speed of migrating Robin is 12 m/s relative to air

so we can say that

$\vec v_{ra} = 12 m/s$ North

so it will be

Let North direction is along Y axis and East direction is along X axis

$\vec v_{ra} = 12\hat j$

also it is given that speed of air is 6.7 m/s relative to ground

$\vec v_a = 6.7 \hat i$

now as we know by the concept of relative motion

$\vec v_{ab} = \vec v_a - \vec v_b$

$\vec v_{ra} = \vec v_r - \vec v_a$

now by rearranging the terms

$\vec v_r = \vec v_{ra} + \vec v_a$

$\vec v_r = 12 \hat j + 6.7 \hat i$

now we need to find the speed of Robin which means we need to find the magnitude of its velocity which we found above

So here we will say

$v_r = \sqrt{12^2 + 6.7^2}$

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3 years ago

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Levart [38]

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2 years ago

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